fuction

[snowpunk]

Allora....

questo è il codice:

<table width="135" border="0" cellpadding="0" cellspacing="5" class="bordotabellacontorno">
  <tr> 
    <td align="center" valign="top" bgcolor="#336666" class="testonews">+ Pi&ugrave; visti +</td>
  </tr>
  <tr> 
    <td align="center" valign="top">
    
<?php 

include("php/config.inc.php");

// immagine
function iconfuv($idicon) {
$query34 = "SELECT id FROM software_icon WHERE id=$idicon LIMIT 1";
$result34 = mysql_query($query34, $db);

while ($row34 = mysql_fetch_array($result34))
{
$icosino = $row34[id];     
}
if($icosino == ""){$icon = "app.png";}     
else{$icon = "vedi.php?id=$idicon";}

return ($icon);
}



$query = "SELECT id, nome, versione, data_mod, c_down, c_visit FROM software ORDER BY c_visit DESC LIMIT 5";
$result = mysql_query($query, $db);
     
$i_icon_c_visit = "0";
while ($row = mysql_fetch_array($result))
{

if($i_icon_c_visit == "0"){

$fuicon = iconfuv($row[id]);

echo"<a href=\"software.php/id/$row[id]\" class=\"testohome\"><img src=\"php/bsoftware/$fuicon\" width=\"80\" height=\"80\"></a>
";
echo"<a href=\"software.php/id/$row[id]\" class=\"testohome\">$row[nome] $row[versione]</a><hr>";
echo"</td></tr>";
}
else{
echo"<tr>";
echo"<td align=\"left\" valign=\"middle\">";
echo"<a href=\"software.php/id/$row[id]\" class=\"testonews\">$row[nome] $row[versione]</a>";
echo"</td></tr>";
}

$i_icon_c_visit ++;

}

mysql_close($db);

?>
    
    
</table>

Mi da questo errore:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /pvisto.php on line 15

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /pvisto.php on line 17 

come mai??